Answer:
{v1,v2,v3} is a basis of given subspace.
v4 = 18/13 v1 + 10/13 v2 + 1/13 v3
v5 = -1/13 v1 -2/13 v2 + 5/13 v3
Step-by-step explanation:
We are done with 3 linearly independent vectors, so either we pick 3 and then prove that they are linearly independent, or we discard 2 of the 5 vectors.
We can start picking v2 = (1,2,2), becuase it starts with a 1, and then we pick v1 = (0,1,3) because it starts with a 0. Since v1 is not 0, then v1 and v2 are linearly independent.
Lets try to complete the basis with v3. For that, we have to show that there exists no linear combination v3 + a * v2 + b * v1 = (0,0,0), for certain real numbers a and b. If that were the case, then
(0,0,0) = (3,1,4) + a*(1,2,2) + b*(0,1,3) = (3 + a, 1+2a+b, 4+2a+3b)
Therefore 3+a = 0, thus a = -3
and 1+2a+b = 1-6+b = -5+b = 0, so b = 5. However
4+2*-3+3*5 = 13 is different from 0. This shows that (v1,v2,v3) are linearly independent and, as a result, they form a basis of R^3 (and in particular, they form a basis of the subspace spanned by {v1,v2,v3,v4,v5})
Also, recall from the previous computation that 5v1-3v2+v3 = (0,0,13), hence
5/13 v1 - 3/13 v2 + 1/13 v3 = (0,0,1)
Now, lets obtain v4 and v5
Note that if we sum v1 and v2 we obtain v1+v2 = (0,1,3)+(1,2,2) = (1,3,5). So we only need (0,0,1) to obtain v4. We replace (0,0,1) with the expression given above and we obtain that
v4 = v1+v2+5/13v1-3/13 v2 + 1/13 v3 = 18/13 v1 + 10/13 v2 + 1/13 v3
On the other hand, if we take v2 and substract from it v1 twice we obtain
v2- 2v1 = (1,2,2) - 2*(0,1,3) = (1,0,-4)
So we only need to add (0,0,5) = 5*(0,0,1) to obtain v5, thus
v5 = v2 - 2* v1 + 5*(5/13 v1 - 3/13 v2 + 1/13 v3) = -1/13 v1 -2/13 v2 + 5/13 v3
